First slide

Theory of equations

Question

There is only one real value of a' for which the quadratic equation ${ax}^{2} + (a + 3) x + a - 3 = 0$ has two positive integral solutions. The solutions is

Moderate

A

9
B

8
C

6
D

12

Solution

Sum of the roots $= - \frac{(a + 3)}{a} = I^{+}$
$a = (- \frac{3}{I^{+} + 1}) - - - - - (i)$
Product of the roots $= αβ = \frac{a - 3}{a} = I^{+} + 2 - - - - (ii)$
and
$\begin{aligned} D & = (a + 3)^{2} - 4 a (a + 3) \\ = \frac{9}{{(I^{+} + 1)}^{2}} {{(I^{+} - 2)}^{2} - 12} \end{aligned}$
D must be perfect square, then I⁺=6
From Eq. (i),
Product of the roots $= I^{+} + 2 = 6 + 2 = 8$

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